Why does this loop run infinite times? Comma separated conditions in for loop

Question

Program code :

int main() 
{ 
    int i; 
    for (i = 0; i < 0, 5; i++) 
        printf("%d ", i); 
    return 0; 
}

The loop above is executed infinite times.

What does i<0,5 mean, and how it is evaluated?

Subexpression `i < 0` will never be the result of expression `(i < 0), 5`. The result of expression `(i < 0), 5` will always be exactly 5, and never anything that is not 5. Value of 5 is considered as `true` so the `for` loop runs forever, because the condition will be `true` forever. — Dialecticus, Feb 01 '20 at 10:11
As illustrated by OPs comments on the answer by Paul Evans, explaining the comma operator is not sufficient to answer this question. It is hence not a duplicate as proposed. — Yunnosch, Feb 01 '20 at 10:12

midor · Accepted Answer · 2020-02-02T10:47:56.563

5

Based on operator precedence, this is interpreted as (i < 0), 5. The comma operator evaluates all statements, but discards their values except the last. So for practical purposes the loop reads as

for (int i = 0; 5; ++i) {...}

Because a non-zero value is interpreted as true in C/C++, this is equivalent to:

for (int i = 0; true; ++i) {...}

which is an infinite loop.

edited Feb 02 '20 at 10:47

answered Feb 01 '20 at 10:07

midor

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Nice explanation, bridging the gap between "How does comma operator work?" and "Why is this infinite?". – Yunnosch Feb 01 '20 at 10:13

Paul Evans · Answer 2 · 2020-02-01T10:08:38.207

2

why i<0,5 how it is evaluated?

Because the value of a comma expression is its last value. The values in this comma expression arei < 0 and 5. This first is evaluated to false and thrown away! The second (and last) is 5 which is true. That is used as the for loop condition expression.

A for loop runs until its condition expression is false, so this loop run forever.

edited Feb 01 '20 at 10:08

answered Feb 01 '20 at 10:01

Paul Evans

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then why it runs infine time – avanish Feb 01 '20 at 10:02
@avanish Added that to answer. – Paul Evans Feb 01 '20 at 10:03
@ Paul Evans but value of i is also increasing so when value of i=5 then it should stop – avanish Feb 01 '20 at 10:04
@avanish Added more details. – Paul Evans Feb 01 '20 at 10:09

0___________ · Answer 3 · 2020-02-01T11:18:43.350

0

It is very easy to check yourself. Try:

int main() 
{ 
    int i, y; 
    for (i = 0; i < 0, 5; i++) 
    {
        y = (i < 0, 5);
        printf("%d ", y); 
    }
    return 0; 
}

And you know that second expression in your list is the value of the comma expression.

edited Feb 01 '20 at 11:18

answered Feb 01 '20 at 10:22

0___________

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score 0 · Answer 4 · answered Feb 01 '20 at 10:32

In the condition of the loop there is used the comma operator

for (i = 0; i < 0, 5; i++) 
    printf("%d ", i);

From the C Standard (6.5.17 Comma operator)

2 The left operand of a comma operator is evaluated as a void expression; there is a sequence point between its evaluation and that of the right operand. Then the right operand is evaluated; the result has its type and value

So this expression with the comma operator

i < 0, 5

has two operands and can be rewritten for clarity like

( i < 0 ), ( 5 )

So the value of the expression according to the quote from the C Standard is the value of the second operand that is 5.

As 5 is not equal to 0 then the condition always evaluates to logical true and you have an infinite loop.

From the C Standard (6.8.5 Iteration statements)

4 An iteration statement causes a statement called the loop body to be executed repeatedly until the controlling expression compares equal to 0. The repetition occurs regardless of whether the loop body is entered from the iteration statement or by a jump.155

Why does this loop run infinite times? Comma separated conditions in for loop

4 Answers4